Get File Name

[Guest]

I need to put a button on a form that will open a File Lookup Window so the
use can open a file on another drive.

My application is in Access 2000.
I found the following code in Access 2003 and I tried using it in 2000 and I
get the error Method or data member not found.

Here is the two step process I'm trying to use:

Private Sub AddDoc_Click()
getFileName
End Sub

Sub getFileName()
' Displays the Office File Open dialog to choose a file name
' for the current employee record. If the user selects a file
' display it in the image control.
Dim fileName As String
Dim result As Integer
With Application.FileDialog(msoFileDialogFilePicker)
.Title = "Select Employee Picture"
.Filters.Add "All Files", "*.*"
.Filters.Add "JPEGs", "*.jpg"
.Filters.Add "Bitmaps", "*.bmp"
.FilterIndex = 3
.AllowMultiSelect = False
.InitialFileName = CurrentProject.Path
result = .Show
If (result <> 0) Then
fileName = Trim(.SelectedItems.Item(1))
Me![ImagePath].Visible = True
Me![ImagePath].SetFocus
Me![ImagePath].Text = fileName
Me![FirstName].SetFocus
Me![ImagePath].Visible = False
End If
End With
End Sub

 

[Guest]

The error is on
..FileDialog

 

[Alex Dybenko]

hi,
you can use this API functions:
http://www.mvps.org/access/api/api0001.htm

 

[Alex Dybenko]

hi,
you can use this API functions:
http://www.mvps.org/access/api/api0001.htm

--
Best regards,
___________
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com

 

[Guest]

When I try this it returns an error on the "ahtAddFilterItem"
Sub or Function not defined.

When I dropdown the available references there is nothing like it - do you
know which reference to check in Access 2000.

Thanks

hi,
you can use this API functions:
http://www.mvps.org/access/api/api0001.htm

--
Best regards,
___________
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com

I need to put a button on a form that will open a File Lookup Window so the
use can open a file on another drive.

My application is in Access 2000.
I found the following code in Access 2003 and I tried using it in 2000 and
I
get the error Method or data member not found.

Here is the two step process I'm trying to use:

Private Sub AddDoc_Click()
getFileName
End Sub

Sub getFileName()
' Displays the Office File Open dialog to choose a file name
' for the current employee record. If the user selects a file
' display it in the image control.
Dim fileName As String
Dim result As Integer
With Application.FileDialog(msoFileDialogFilePicker)
.Title = "Select Employee Picture"
.Filters.Add "All Files", "*.*"
.Filters.Add "JPEGs", "*.jpg"
.Filters.Add "Bitmaps", "*.bmp"
.FilterIndex = 3
.AllowMultiSelect = False
.InitialFileName = CurrentProject.Path
result = .Show
If (result <> 0) Then
fileName = Trim(.SelectedItems.Item(1))
Me![ImagePath].Visible = True
Me![ImagePath].SetFocus
Me![ImagePath].Text = fileName
Me![FirstName].SetFocus
Me![ImagePath].Visible = False
End If
End With
End Sub

 

[Douglas J. Steele]

Sounds as though you didn't copy all of the code from the cited page.

You need to copy everything between the Code Start and Code Ends lines into
a new module (not a class module nor a module associated with a form).

--
Doug Steele, Microsoft Access MVP

(no e-mails, please!)

When I try this it returns an error on the "ahtAddFilterItem"
Sub or Function not defined.

When I dropdown the available references there is nothing like it - do you
know which reference to check in Access 2000.

Thanks

hi,
you can use this API functions:
http://www.mvps.org/access/api/api0001.htm

--
Best regards,
___________
Alex Dybenko (MVP)
http://alexdyb.blogspot.com
http://www.PointLtd.com

I need to put a button on a form that will open a File Lookup Window so
the
use can open a file on another drive.

My application is in Access 2000.
I found the following code in Access 2003 and I tried using it in 2000
and
I
get the error Method or data member not found.

Here is the two step process I'm trying to use:

Private Sub AddDoc_Click()
getFileName
End Sub

Sub getFileName()
' Displays the Office File Open dialog to choose a file name
' for the current employee record. If the user selects a file
' display it in the image control.
Dim fileName As String
Dim result As Integer
With Application.FileDialog(msoFileDialogFilePicker)
.Title = "Select Employee Picture"
.Filters.Add "All Files", "*.*"
.Filters.Add "JPEGs", "*.jpg"
.Filters.Add "Bitmaps", "*.bmp"
.FilterIndex = 3
.AllowMultiSelect = False
.InitialFileName = CurrentProject.Path
result = .Show
If (result <> 0) Then
fileName = Trim(.SelectedItems.Item(1))
Me![ImagePath].Visible = True
Me![ImagePath].SetFocus
Me![ImagePath].Text = fileName
Me![FirstName].SetFocus
Me![ImagePath].Visible = False
End If
End With
End Sub

 

[Guest]

Did you add 'Microsoft Office 11.0 Object Library' to your references?